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A 1.25-\muCi alpha source gives off alpha particles each with a kinetic energy of \(2.78 \mathrm{MeV}\). At what rate (in watts) is kinetic energy being produced?

Short Answer

Expert verified
Approx. \(2.06 \times 10^{-8} \, \mathrm{W}\).

Step by step solution

01

Convert Microcuries to Decays per Second

To find the decay rate, we first need to convert the activity of the source from microcuries (\( \mu \mathrm{Ci} \)) to decays per second (\( \mathrm{Bq} \)). Recall that \[1 \, \mathrm{Ci} = 3.7 \times 10^{10} \, \mathrm{Bq}\]Therefore,\[1.25 \, \mu \mathrm{Ci} = 1.25 \times 10^{-6} \, \mathrm{Ci} = 1.25 \times 10^{-6} \times 3.7 \times 10^{10} \, \mathrm{Bq} = 46,250 \, \mathrm{Bq}\]This means the decay rate is 46,250 decays per second.
02

Convert Kinetic Energy from MeV to Joules

Next, we need to convert the kinetic energy of the alpha particles from MeV to Joules. Recall that \[1 \, \mathrm{eV} = 1.602 \times 10^{-19} \, \mathrm{J}\]Thus,\[1 \, \mathrm{MeV} = 1 \times 10^6 \times 1.602 \times 10^{-19} \, \mathrm{J} = 1.602 \times 10^{-13} \, \mathrm{J}\]Therefore,\[2.78 \, \mathrm{MeV} = 2.78 \times 1.602 \times 10^{-13} \, \mathrm{J} = 4.45356 \times 10^{-13} \, \mathrm{J}\]
03

Calculate Power in Watts

Now, we can calculate the power in watts by multiplying the decay rate by the kinetic energy of each alpha particle. Using the values from the previous steps:\[\text{Power} = \text{Decay Rate} \times \text{Energy per Decay}\]\[= 46,250 \, \text{decays/sec} \times 4.45356 \times 10^{-13} \, \text{J/decay}\]\[ = 2.060805 \times 10^{-8} \, \mathrm{W}\]Thus, the power is approximately \(2.06 \times 10^{-8} \, \mathrm{W}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactivity
Radioactivity is a process by which unstable atomic nuclei lose energy by emitting radiation, such as alpha particles, beta particles, or gamma rays. This process is important in nuclear physics because it explains how elements transform into other elements over time due to natural decay.

Alpha particles, which consist of two protons and two neutrons, are a common type of radiation emitted in this process. They are positively charged and relatively heavy compared to other forms of radiation like beta particles or gamma rays.

When studying an alpha source, such as the one described in the exercise, it is useful to measure its activity in units like Curie (Ci) or Becquerel (Bq). The activity indicates how many nuclei decay per unit of time. For instance, the source with an activity of 1.25 microcuries, which is equivalent to 46,250 decays per second, shows a relatively high rate of radiation emission.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. In the context of nuclear physics and radioactivity, particles such as alpha particles have kinetic energy when they are emitted from the nucleus during the decay process.

In the exercise, each emitted alpha particle has a kinetic energy of 2.78 MeV. It is often necessary to convert this energy into more familiar units such as joules to perform calculations within the International System of Units (SI).
  • 1 electronvolt (eV) is equal to 1.602 x 10^-19 joules.
  • Therefore, 1 MeV, which is one million electronvolts, equals 1.602 x 10^-13 joules.
By understanding the conversion of MeV to joules, we can calculate how much kinetic energy each alpha particle carries, which helps us assess the power output from radioactive decay.
Energy Conversion
Energy conversion is a crucial aspect of physics that examines how energy changes from one form to another. In nuclear physics, particularly in the study of radioactivity, the conversion involves the transformation of nuclear energy into kinetic energy and eventually into other forms of energy, such as thermal energy.

For the alpha source in the problem, the energy conversion process specifically translates the decay of alpha particles into kinetic energy, which is then measured in joules.

To compute the rate at which this energy is produced, we multiply the decay rate by the energy of each particle. Thus, by knowing both the decay rate and the kinetic energy of each decay, one can easily calculate how much energy is being converted perpetually, as shown in the exercise final result of 2.06 x 10^-8 watts.
Decay Rate
Decay rate is a measure of how quickly radioactive atoms in a sample disintegrate over time. It's typically expressed in decays per second using the unit Becquerel (Bq), which signifies one decay event per second.

In our exercise, the alpha source with an activity of 1.25 microcuries gives a decay rate of 46,250 Bq. This means that every second, 46,250 alpha decay events occur.
  • Decay rate helps determine how fast a radioactive substance changes.
  • It also directly influences the overall power output when combined with the kinetic energy per decay event.
By understanding the decay rate, scientists can predict the longevity and energy output of radioactive materials, which are key elements in applications ranging from medical treatments to energy generation.

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Most popular questions from this chapter

\({ }_{1}^{3} \mathrm{H}\) (tritium) can be produced in water surrounding a strong source of neutrons, such as that occuring in nuclear reactors. One of the ways tritium can form is via neutron capture by deuterium. (a) Write down the equation for this capture reaction. (b) Tritium has a half-life of 12.33 years. What percentage of a sample containing \({ }_{1}^{3} \mathrm{H}\) will remain after exactly 6 years? (c) Determine the gamma-ray energy emitted during the capture (assuming the tritium ends up in its ground state and the incoming neutron kinetic energy is negligible). (d) Write down the reaction for the subsequent beta decay of the tritium and determine the stable daughter identity. (e) If all the energy released in the beta decay went into the beta particle, determine its energy.

The half-life of a radioactive isotope is known to be exactly \(1 \mathrm{~h}\). (a) What fraction of a sample would be left after exactly \(3 \mathrm{~h}:(1)\) one-third, (2) one-eighth, or (3) oneninth? (b) What fraction of a sample would be left after exactly 1 day?

A sample of technetium- 104 , which has a half-lfe of \(18.0 \mathrm{~min}\), has an initial activity of \(10.0 \mathrm{mCi}\). Determine the activity of the sample after exactly 1 h has elapsed.

A cancer treatment called the gamma knife (see Insight 29.1, Biological and Medical Application of Radiation) uses focused \({ }^{60}\) Co sources to treat tumors. Each \({ }^{60}\) Co nucleus emits two gamma rays, of energy \(1.33 \mathrm{MeV}\) and \(1.17 \mathrm{MeV}\), in quick succession. Assume that \(50.0 \%\) of the total gamma-ray energy is absorbed by a tumor. Further assume that the total activity of the \({ }^{60}\) Co sources is \(1.00 \mathrm{mCi}\), the tumor's mass is \(0.100 \mathrm{~kg}\), and the patient is exposed to the gamma radiation for an hour. Determine the effective radiation dose received by the tumor. (Since the \({ }^{60}\) Co half-life is 5.3 years, changes in its activity during treatment are negligible.)

At present, a radioactive beta source with a long halflife has an activity of \(20 \mathrm{mCi}\). (a) What is the present decay rate in decays per second? (b) Assuming that one beta particle is emitted per decay, how many are currently emitted per minute?

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